Prove the statement wrong, fix it, then prove the new statement:

If {$a_n$} and {$b_n$} are increasing, then {$a_n b_n$} is increasing.

I think I'm headed in the right direction with this but I'm not sure.
What I've tried:
Consider the increasing sequence $\{\frac{-1}{n}\}$. Then $\{(\frac{-1}{n})(\frac{-1}{n})\}=\{\frac1{n^2}\}$. Since $\{\frac1{n^2}\}$ is decreasing, it makes a counterexample to the statement. The statement can then be amended:
If $\{a_n\}$ and $\{b_n\}$ are increasing and $a_n>0 $and $b_n>0$ for every $n\ge1$, then $\{a_n b_n\}$ is increasing.


Since $\{a_n\}$ and $\{b_n\}$ are increasing, we know that $a_{n+1}>a_n$ and $b_{n+1}>b_n$. Then consider two cases for the value of $b_n$:

(i) If $b_n>0$, then $(a_n)(b_n)<(a_{n+1})(b_{n+1}) \implies a_nb_n<a_{n+1}b_{n+1}$ and $\{a_nb_n\}$ is increasing.

(ii) If $b_n<0$, then $(a_n)(b_n)<(a_{n+1})(b_{n+1}) \implies a_nb_n>a_{n+1}b_{n+1}$ and $\{a_nb_n\}$ is decreasing.

Therefore $\{a_nb_n\}$ is increasing when $a_n>0$ and $b_n>0$ for every $n\ge1$.

  • 1
    $\begingroup$ What is your question? Your solution looks good for counter-example and how to fix the claim, and you haven't asked for any specific kind of responses. $\endgroup$ May 30, 2015 at 19:01
  • $\begingroup$ I wanted to know if I was headed in the right direction with my proof. I know it's not the most complete, but do I have the right ideas? $\endgroup$ May 30, 2015 at 19:09
  • 4
    $\begingroup$ Since you are assuming $b_n\gt 0$ there is no case $b_n\lt 0$.. You could write $a_nb_n\lt a_{n+1}b_n\lt a_{n+1}b_{n+1}$ $\endgroup$ May 30, 2015 at 19:27


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